A DC ammeter and n AC thermal ammeter are connected to a circuit in series. When a direct current is passed through the cirucit, the DC ammeter shows. `i_(1)=6A`. When a sinusoidal alternating current flows through the cirucit, the DC ammeter shows `i_(2)=8A`. If both `i_(1) and i_(2)` are passed through the circuit

Options

A. dc=6 A, ac=10A
B. dc=3 A, ac=5A
C. dc=5 A, ac=8A
D. dc=2 A, ac=3A

Answer

A

Explanation

Resultant current is superposition of tow currents, i.e.
I(instantaneous total current ) `=6+(I_0) sin omega t`
dc ammeter will read average value
`=bar(6+(I_0)sin omega t) = 6` `(: Bar((I_0) sin omega t =0))`
ac ammeter will read
`=sqrt((bar(6+(I_0)sin omega t^(2)))`
`=sqrt(bar(36+12 I_(0) sin omega t +I_(0)^(2) sin ^(2)omega t))` (`because bar(I_(0)si omega t =0))`
since `bar(sin^(2) ometa t)=1/2 and (I_(rms))=8=(I_0)/(sqrt(2) implies (I_0)= 8 sqrt(2)A`
Therefore ac reading
`=sqrt(36 + (I_(0)^(2))/(2)) = sqrt(36 + 64) = 10 A`.